/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 18 A \(250 \mathrm{~g}\) block slid... [FREE SOLUTION] | 91Ó°ÊÓ

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A \(250 \mathrm{~g}\) block slides on a rough horizontal table. Find the work done by the frictional force in bringing the block to rest if it is initially moving at a speed of 40 \(\mathrm{cm} / \mathrm{s}\). If the friction coefficient between the table and the block is \(0^{\circ} 1\), how far does the block move before coming to rest"?

Short Answer

Expert verified
The work done by friction is \(-0.02 \text{ J}\), and the block moves approximately \( 0.0816 \text{ m} \) before stopping.

Step by step solution

01

Convert Units

First, convert the block's initial speed from centimeters per second to meters per second. Since there are 100 centimeters in a meter, the initial speed is \( 40 \text{ cm/s} = 0.4 \text{ m/s} \).
02

Calculate Initial Kinetic Energy

The initial kinetic energy (KE) of the block is given by the formula \( KE = \frac{1}{2} mv^2 \), where \( m \) is the mass and \( v \) is the velocity. Here, \( m = 0.25 \text{ kg} \) and \( v = 0.4 \text{ m/s} \). Thus, \( KE = \frac{1}{2} \times 0.25 \times (0.4)^2 = 0.02 \text{ J} \).
03

Work Done by Friction

The work done by the frictional force to bring the block to rest is equal to the initial kinetic energy, which is \( 0.02 \text{ J} \). Since work done by friction is negative work, it equals \( -0.02 \text{ J} \).
04

Calculate Frictional Force

The frictional force \( f \) can be found using \( f = \mu N \), where \( \mu \) is the coefficient of friction and \( N \) is the normal force. The normal force for a block on a horizontal table is equal to the weight of the block, \( N = mg \), where \( g = 9.8 \text{ m/s}^2 \). Thus, \( f = 0.1 \times 0.25 \times 9.8 = 0.245 \text{ N} \).
05

Calculate Distance Moved

The work done by friction is also given by \( W = f \times d \), where \( d \) is the distance moved. Rearranging this, \( d = \frac{W}{f} = \frac{0.02}{0.245} \approx 0.0816 \text{ m} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Work Done by Friction
When we talk about the work done by friction, we're discussing how much energy the frictional force takes from a moving object until it stops. In physics, work is calculated as the force applied multiplied by the distance over which the force is applied. However, in friction's case, it always acts in the opposite direction of motion, so the work done is considered negative.

In our exercise, the work done by friction brings the moving block to a stop. The initial kinetic energy of the block is converted into heat energy by the frictional force. Hence, the work done by friction equals the block’s initial kinetic energy, which is \( -0.02 ext{ J} \). The negative sign indicates that friction is taking energy away from the block.
Kinetic Energy
Kinetic energy is the energy that an object possesses due to its motion. It's calculated using the formula \( KE = \frac{1}{2} mv^2 \), where \( m \) is the mass and \( v \) is the velocity of the object. In simpler terms, the faster an object moves, or the heavier it is, the more kinetic energy it has.

In our case, the block starts with kinetic energy because it's moving at a speed of \( 0.4 ext{ m/s} \). With a mass of \( 0.25 ext{ kg} \), its initial kinetic energy can be calculated to be \( 0.02 ext{ J} \). The frictional force then works to convert this energy into heat, stopping the block in the process.
Frictional Force
Frictional force is the resistance that one surface or object encounters when moving over another. In our exercise, this force acts between the block and the table surface, opposing the block's motion.

Frictional force is a crucial factor because it determines how quickly the block stops moving. The magnitude of this force can be calculated using \( f = \mu N \), where \( \mu \) is the coefficient of friction, and \( N \) is the normal force, typically equal to the object's weight on a horizontal surface. For our block, the frictional force was \( 0.245 ext{ N} \), which led to the block coming to a stop over a specific distance.
Coefficient of Friction
The coefficient of friction is a dimensionless number that represents the degree of interaction between two surfaces. A higher coefficient means more frictional force for a given normal force. In our problem, the coefficient of friction between the block and the table is \( 0.1 \).

This means that the surfaces experience a moderate degree of friction, not too high and not too low. It's used to calculate the frictional force and helps predict how far the block will slide before stopping. This calculation involves plugging the coefficient into the formula for frictional force \( f = \mu N \), turning kinetic energy into friction work until the block halts.

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Most popular questions from this chapter

A block of mass \(5^{\circ} 0 \mathrm{~kg}\) slides down an incline of inclination \(30^{\circ}\) and length \(10 \mathrm{~m}\). Find the work done by the force of gravity.

A block of mass \(2 \cdot 0 \mathrm{~kg}\) kept at rest on an inclined plane of inclination \(37^{\circ}\) is pulled up the plane by applying a constant force of \(20 \mathrm{~N}\) parallel to the incline. The force acts for one second. (a) Show that the work done by the applied force does not exceed \(40 \mathrm{~J} .\) (b) Find the work done by the force of gravity in that one second if the work done by the applied force is \(40 \mathrm{~J} .(\mathrm{c})\) Find the kinetic energy of the block at the instant the force ceases to act. Take \(g=10 \mathrm{~m} / \mathrm{s}^{2}\).

The mass of cyclist together with the bike is \(90 \mathrm{~kg}\). Calculate the increase in kinetic energy if the speed increases from \(6 \cdot 0 \mathrm{~km} / \mathrm{h}\) to \(12 \mathrm{~km} / \mathrm{h}\).

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